Part I

Chapter I 微积分的概念

1 函数与极限

1.1

$\lim\limits_{n\to\infty}a_n=A$ ¹.

$\lim\limits_{x\to a}f(x)=A$ ² $a$ 可以是有限数, 也可以是无穷大.

1.2

$f(x)$ $t=t_0$ $\lim\limits_{t\to t_0}f(t)=f(t_0)$ 成立, 否则称为间断.

2 定积分

2.1

$\{\Delta x_i\}$ $y=x^m$ $m$ $m\neq-1$ $x$ $x=a$ $x=b$ $\frac{b^{m+1}-a^{m+1}}{m+1}$ .

2.3

$\int_1^x\frac{1}{s}{\mathrm{d}s}=\ln x$ $e$ $\ln e=1$
$\ln{x_1 x_2}=\ln x_1+\ln x_2$ ³ $\ln x$ $e$ $\ln x=\int_1^x\frac{1}{s}{\mathrm{d}s}=\log_e x$ .

3 微商与微分

3.3

$\lim\limits_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x)}{\Delta x}$ $f(x)$ $x_0$ $f(x)$ $x_0$ $f'(x_0)$ $\frac{\mathrm{d}y}{\mathrm{d}x}\big|_{x=x_0}$ .

$y=f(x)$ $x=x_0$ 处可微, 则一定在这一点连续⁴.

$\lim\limits_{n\to\infty}{(1+\frac{1}{n})^n}=e$ ⁵.

3.4

$y=f(x)$ $x$ $\mathrm{d}y=f'(x)\Delta x$
$\mathrm{d}y$ $\Delta y$ ⁶.

$\mathrm{d}y=f'(x)\Delta x$ $\mathrm{d}y=f'(x)\mathrm{d}x$ ⁷,
这就是微商这个名称的由来.

3.5

Fermat 定理⁸
Rolle 定理⁹

微分中值定理
Lagrange 中值定理 (微分中值定理)
Cauchy 中值定理¹⁰

Chapter II 微积分的运算

1 微分法

1.1

$y=f(u),u=\varphi (x)$ $\frac{\mathrm{d}y}{\mathrm{d}x}=f'(u)\varphi'(x)$ ¹¹.

$x=g(y)$ $y=f(x)$ $f'(x)\neq0$ $g'(y)=\frac{1}{f'(x)}$ ¹².

$(\tan x)'=\sec^2x$
$(\cot x)'=-\csc^2x$
$(\sec x)'=\sec x\tan x$
$(\csc x)'=-\csc x\cot x$
$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$ ¹³
$(\arccos x)'=\frac{-1}{\sqrt{1-x^2}}$
$(\arctan x)'=\frac{1}{1+x^2}$
$(\mathrm{arccot}\,x)'=\frac{-1}{1+x^2}$
…more¹⁴

一阶微分形式的不变性¹⁵

1.2

$f^{(n-1)}(x)$ $f(x)$ $n$ $f^{(n)}(x)$ $\frac{\mathrm{d}^nf(x)}{\mathrm{d}x^n}$ ¹⁶.

Leibnitz 公式¹⁷

$\left\{\begin{array}\\x=\varphi(t)\\y=\psi(t)\end{array}\right.$ 所确定的函数的微分法¹⁸

2 积分法

2.1

$\int\frac{1}{x^2-a^2}\mathrm{d}x=\frac{1}{2a}\ln|\frac{x-a}{x+a}|$
$\int\sec x\mathrm{d}x=\ln|\sec x+\tan x|$
$\int\sqrt{a^2-x^2}\mathrm{d}x=\frac{a^2}{2}\arcsin\frac{x}{a}+\frac{x}{2}\sqrt{a^2-x^2}$
$\int\sqrt{x^2+a^2}\mathrm{d}x=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln(x+\sqrt{x^2+a^2})$
$\int\frac{\mathrm{d}x}{\sqrt{x^2\pm1}}=\ln(x+\sqrt{1\pm x^2})$
$\int\arctan x\mathrm{d}x=x\arctan x-\frac{1}{2}\ln(x^2+1)$
$\int e^{ax}\cos bx\mathrm{d}x=\frac{b\sin bx+a\cos bx}{a^2+b^2}e^{ax}$
$\int e^{ax}\sin bx\mathrm{d}x=\frac{a\sin bx-b\cos bx}{a^2+b^2}e^{ax}$
$J_n=\int\frac{\mathrm{d}x}{(x^2+a^2)^n},J_1=\frac{1}{a}\arctan\frac{x}{a},J_{n+1}=\frac{1}{2na^2}\frac{x}{(x^2+a^2)^n}+\frac{2n-1}{2n}\frac{1}{a^2}J_n$
$\int_0^{\frac{\pi}{2}}\sin^mx\mathrm{d}x=\int_0^{\frac{\pi}{2}}\cos^mx\mathrm{d}x=\left\{\begin{array}\\\frac{(m-1)!!}{m!!}\frac{\pi}{2}&(\text{even}\,m)\\\frac{(m-1)!!}{m!!}&(\text{odd}\,m)\end{array}\right.$

$\int\frac{mx+n}{(x^2+px+q)^k}\mathrm{d}x=\int\frac{mx+n}{((x+\xi)^2+\eta)^k}\mathrm{d}x=a\int\frac{t\mathrm{d}t}{(t^2+\eta)^k}+b\int\frac{\mathrm{d}t}{(t^2+\eta)^k}$ , 查表易得.

2.2

换元法
$\int_{a}^{b} f (x) d x = \int_{φ^{- 1} (a)}^{φ^{- 1} (b)} f [φ (t)] φ^{'} (t) d t$
分部积分法

Chapter III 微积分的一些应用

2 曲线的描绘

2.5

$y=f(x)$ $k=\lim\limits_{\Delta s\to0}\frac{\Delta\varphi}{\Delta s}=\frac{|y''|}{(1+y'^2)^{3/2}}$
$\left\{\begin{array}\\x=\varphi(t)\\y=\psi(t)\end{array}\right.$ $k=\frac{|\psi''\varphi'-\psi'\varphi''|}{[\varphi'^2+\psi'^2]^{3/2}}$ .

3 Taylor 展开与极值问题

3.1

$f(x)$ $x=x_0$ 的泰勒展开式¹⁹为

\begin{aligned} f (x) = & f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + \frac{f^{″} (x_{0})}{2!} (x - x_{0})^{2} \\ + \dots + \frac{f^{(n)} (x_{0})}{n!} (x - x_{0})^{n} + \frac{f^{(n + 1)} (ξ)}{(n + 1)!} (x - x_{0})^{n + 1}, \end{aligned}

$\xi=x_0+\theta(x-x_0),0<\theta<1$ . (显然, Lagrange 中值定理只是它的一个最简单的特例)

$e^x=\sum\frac{x^n}{n!}$
$\sin x\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}$

L’ Hospital 法则²⁰

Chapter IV 常微分方程

1 一阶微分方程

1.1

$y$ $n$ 阶线性微分方程的一般形式²¹

1.2

分离变量法²²
$y'=f(\frac{y}{x})$ 的方程经过变换可以变成可分离变量的²³

1.3

$\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)$ 的解²⁴为

e^{- \int P (x) d x} \int Q (x) e^{\int P (x) d x} d x

$\int P(x)\mathrm{d}x$ $P(x)$ 的一个原函数.

常数变易法²⁵

2 二阶微分方程

2.1

$y''=f(x,y')$ ²⁶ $y''=f(y,y')$ ²⁷

2.2

二阶线性常微分方程的一般形状是

y^{″} + p (x) y^{'} + q (x) y = f (x) (2.2 .1) .

$f(x)\equiv0$ , 则方程称为齐次的, 此时方程的初值问题是指

{\begin{matrix} y^{″} + p (x) y^{'} + q (x) y = 0 & (2.2 .2) \\ y (x_{0}) = α, y^{'} (x_{0}) = β \end{matrix},

它的解是存在且唯一的 (假定已证), 由此可以得到 (2.2.2) 的解的结构定理: 任意解可表成一对线性无关解的线性组合. 下面来证明这条结构定理:

$\varphi(x),\psi(x)$ 线性相关, 则它们的 Wronski 行列式²⁸
$W (φ, ψ) = | \begin{matrix} φ (x) & ψ (x) \\ φ^{'} (x) & ψ^{'} (x) \end{matrix} | \equiv 0,$
$W(\varphi,\psi)\equiv0$ ²⁹ $\varphi(x),\psi(x)$ $x_0$ $W(\varphi(x_0),\psi(x_0))=0\Leftrightarrow$ $\varphi(x)$ $\psi(x)$ 线性相关³⁰.
$y_1(x),y_2(x)$ $W(y_1(x),y_2(x))=W(x)$ ³¹ $W(x)=e^{-\int P(x)\mathrm{d}x}$ $W(x)$ 在某点为零, 与处处为零是等价的.
$y_1(x),y_2(x)$ $x_0$ $W(x_0)=\left|\begin{array}\\y_1(x_0)&y_2(x_0)\\y_1'(x_0)&y_2'(x_0)\end{array}\right|\neq0$ .
$y(x)$ $W(x_0)\neq0$ , 一定存在
$\begin{array}{r} C_{1} y_{1} (x_{0}) + C_{2} y_{2} (x_{0}) = y (x_{0}), \\ C_{1} y_{2}^{'} (x_{0}) + C_{2} y_{2}^{'} (x_{0}) = y^{'} (x_{0}) . \end{array}$
$y(x)$ $C_1y_1(x)+C_2y_2(x)$ $y(x)=C_1y_1(x)+C_2y_2(x)$ 成立. 这就证明了齐次二阶线性常微分方程的解的结构定理.

$y_1(x)(\not\equiv0)$
$y_2=y_1\cdot\int\frac{1}{y_1^2}e^{-\int P(x)\mathrm{d}x}\mathrm{d}x$ ³².

可以用常数变易法求出非齐次方程 (2.2.1) 的特解³³

y^{*} (x) = - y_{1} \int \frac{y_{2} f d x}{W (y_{1}, y_{2})} + y_{2} \int \frac{y_{1} f d x}{W (y_{1}, y_{2})},

与 (2.2.2) 的通解相加, 得出 (2.2.1) 的通解.

2.3

二阶常系数线性微分方程形如

y^{″} + a y^{'} + b y = f (x),

求解它的问题归结于求解它相应的齐次方程. 这是容易做到的, 如下:

$y''+ay'+by=0$ $y=e^{\lambda x}$ 形式的解, 回代这个形式, 得出特征方程
$λ^{2} + a λ + b = 0.$
$\lambda_1,\lambda_2$ $y=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$ .
$\lambda=-\frac{a}{2}$ $y_1=e^{-\frac{a}{2}x},y_2=xe^{-\frac{a}{2}x}$ ³⁴ $y=e^{-\frac{a}{2}x}(C_1+C_2x)$ .
$\lambda=\alpha\pm\mathrm{i}\beta$ $y=e^{ax}(C_1\cos\beta x+C_2\sin\beta x)$ ³⁵.

Part II

Chapter VI 重积分与偏微商

1 重积分

1.1

$P(x,y)$ $A(a,b)$ $f(x,y)$ $l$ $l$ $f(x,y)$ $P(x,y)$ $A(a,b)$ $\lim\limits_{x\to a\atop y\to b}f(x,y)=l$ $\lim\limits_{P\to A}f(x,y)=l$
$\lim\limits_{(x,y)\to(a,b)}f(x,y)=f(a,b)$ $f(x,y)$ $(a,b)$ 处是连续的.

1.2

$f(x,y)$ $D$ $\varphi(x,y)$ $D$ $D$ $(\xi,\eta)$ $\iint\limits_{D}f(x,y)\varphi(x,y)\mathrm{d}A=f(\xi,\eta)\iint\limits_{D}\varphi(x,y)\mathrm{d}A$ .

1.3

\begin{aligned} \iint_{D} f (x, y) d A = \int_{a}^{b} (\int_{φ_{1} (x)}^{φ_{2} (x)} f (x, y) d y) d x = \int_{a}^{b} d x \int_{φ_{1} (x)}^{φ_{2} (x)} f (x, y) d y = \iint_{D} f (x, y) d x d y . \\ \underset{V}{∭} f (x, y, z) d V = \underset{V}{∭} f (x, y, z) d x d y d z \\ = \iint_{D} d x d y \int_{z_{1} (x, y)}^{z_{2} (x, y)} f (x, y, z) d z = \int_{g}^{h} d z \iint_{D_{z}} f (x, y, z) d x d y . \end{aligned}

2 偏微商

2.1

$\lim\limits_{\Delta x\to0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}=\frac{\partial f}{\partial x}=f'_x(x,y)$ $f(x,y)$ $x$ $\frac{\partial f}{\partial x}\mathrm{d}x$ $f(x,y)$ $x$ $f(x,y)$ $(x,y)$ $\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y$ .

全微分除了有一些与单变量函数的微分法则在形式上一致的性质³⁶, 还有了新的内容. 简单地说, 与 Jacobi 矩阵有关, 并由此可以推出微分的不变性³⁷.

$\frac{\partial }{\partial y}(\frac{\partial f}{\partial x})$ $\frac{\partial^2f}{\partial y\partial x}$ $f''_{xy}$
$\frac{\partial^2f}{\partial x\partial y}$ $\frac{\partial^2f}{\partial y\partial x}$ $(x,y)$ $\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x}$ , 即微商的次序可以交换³⁸.

$z=f(x,y),y=y(x)$ $\frac{\partial z}{\partial x}$ $\frac{\partial f}{\partial x}$ 是不同的.

2.2

$F(x,y,z)=0$ $z$ $x,y$ $z=z(x,y)$ $z$ $x$ $\frac{\partial z}{\partial x}=-F'_x/F'_z$ ³⁹
$\left\{\begin{array}\\F(x,y,\cdots,u,v)=0\\G(x,y,\cdots,u,v)=0\end{array}\right.$ $u,v$ $\frac{\partial u}{\partial x}=-\frac{\partial (F,G)}{\partial (x,v)}/\frac{\partial (F,G)}{\partial (u,v)}$ ⁴⁰.

3 Jacobi 行列式, 面积元素与体积元素

3.1

$u=u(x,y),v=v(x,y)$ $\left[\begin{array}\\\frac{\partial u}{\partial x}&\frac{\partial v}{\partial x}\\\frac{\partial u}{\partial y}&\frac{\partial v}{\partial y}\end{array}\right]$ $u$ $v$ $\frac{\partial(u,v)}{\partial(x,y)}$ . 多余两个变量的函数也可类似地定义 Jacobi 矩阵.

$\left\{\begin{array}\\u=u(x,y)\\v=v(x,y)\end{array}\right.,\left\{\begin{array}\\x=x(s,t)\\y=y(s,t)\end{array}\right.$ ):

$\frac{\partial(u,v)}{\partial(s,t)}=\frac{\partial(u,v)}{\partial(x,y)}\cdot\frac{\partial(x,y)}{\partial(s,t)}$ .
$\frac{\partial(u,v)}{\partial(x,y)}\cdot\frac{\partial(x,y)}{\partial(u,v)}=\frac{\partial(u,v)}{\partial(u,v)}=\left|\begin{array}\\1&0\\0&1\end{array}\right|=1$ $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}$ 相当.
$\frac{\partial(u,v)}{\partial(x,y)}=-\frac{\partial(v,u)}{\partial(x,y)}$ .
$\frac{\partial(u,u)}{\partial(x,y)}=0$ .

3.2

$\left\{\begin{array}\\x=x(u,v)\\y=y(u,v)\end{array}\right.$ $(u,v)$ $xOy$ $\mathrm{d}A=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\mathrm{d}u\mathrm{d}v$ $\iint\limits_Df(x,y)\mathrm{d}A=\iint\limits_\mathscr{D}f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\mathrm{d}u\mathrm{d}v$ .

极坐标的面素⁴¹
球坐标 (空间极坐标) 的体素⁴²
柱坐标的体素⁴³
直角三棱锥的一个可行的变换⁴⁴

$\vec r=\vec r(u,v)$ $\mathrm{d}S=\sqrt{EG-F^2}\mathrm{d}u\mathrm{d}v$ ⁴⁵
$z=z(x,y)$ 的情形则更简单⁴⁶.

Chapter VII 线, 面积分与外微分形式

1 数量场与矢量场

1.1

$u=u(x,y,z),(x,y,z)\in V$ $V$ $M_0$ $l$ $M_0$ $u=u(M)$ $l$ $(\frac{\partial u}{\partial l})_{M_0}=\lim\limits_{M\to M_0\atop M\in l}\frac{u(M)-u(M_0)}{|MM_0|}$ $u$ $M_0$ $(\mathbf{grad}\,u)_{M_0}=(\frac{\partial u}{\partial x}\vec i+\frac{\partial u}{\partial y}\vec j+\frac{\partial u}{\partial z}\vec k)_{M_0}$ ⁴⁷.

$\nabla=\vec i\frac{\partial}{\partial x}+\vec j\frac{\partial}{\partial y}+\vec k\frac{\partial}{\partial z}$ . 以下用此符号描述一些梯度的性质:

$\frac{\partial u}{\partial l}=\nabla u\cdot\boldsymbol l^0$ .
沿梯度方向, 数量场的变化最快, 等值面分布最 “密”.
梯度运算法则:
$\nabla Cu=C\nabla u$
$\nabla(u+v)=\nabla u+\nabla v$
$\nabla uv=u\nabla v+v\nabla u$
$\nabla f(u)=f'(u)\nabla u$ .

1.2

$\vec v=P(x,y,z)\vec i+Q(x,y,z)\vec j+R(x,y,z)\vec k$ , 它的流线族由微分方程组解得⁴⁸.

2 曲线积分

2.1 关于弧长的曲线积分

$L$ $s$ $\left\{\begin{split}x=x(t)\\y=y(t)\\z=z(t)\end{split}\right.,\alpha\leqslant t\leqslant\beta$ $\dot x,\dot y,\dot z$ $L$ $f(M)=f(x,y,z)$ $L$ $f(M)$ $L$ 上的第一种曲线积分存在, 且有

\int_{L} f (M) d s = \int_{α}^{β} f (x, y, z) \sqrt{{\dot{x}}^{2} + {\dot{y}}^{2} + {\dot{z}}^{2}} d t .

$L$ $y=y(x),a\leqslant x\leqslant b$
$\int_Lf(M)\mathrm{d}s=\int_Lf(x,y)\mathrm{d}s=\int_a^bf(x,y(x))\sqrt{1+y'^2}\mathrm{d}x$ .
$L$ $r=r(\theta),\alpha\leqslant\theta\leqslant\beta$
$\int_Lf(M)\mathrm{d}s=\int_Lf(x,y)\mathrm{d}s=\int_\alpha^\beta f(r(\theta)\cos\theta,r(\theta)\sin\theta)\sqrt{r^2(\theta)+r'^2(\theta)}\mathrm{d}\theta$ .

2.3 关于弧长元素投影的积分

$\int_{L_{AB}}f(M)\mathrm{d}x$ ⁴⁹. 以下罗列一些简单性质:

$\int_{L_{AB}}f(M)\mathrm{d}x=-\int_{L_{BA}}f(M)\mathrm{d}x$
$\int_{L_{AB}}f\mathrm{d}x+\int_{L_{BC}}f\mathrm{d}x=\int_{L_{AC}}f\mathrm{d}x$
$L$ $xOy$ $Oy$ $\int_Lf\mathrm{d}x=0$ $L$ $Oxyz$ $yOz$ $L$ $x$ 的第二种曲线积分为零.

2.4

$Oxyz$ $L$ $\left\{\begin{split}x=x(t)\\y=y(t)\\z=z(t)\end{split}\right.,\alpha\leqslant t\leqslant\beta$ $\dot x,\dot y,\dot z$ $A,B$ $A(x(\alpha),y(\alpha),z(\alpha)),B(x(\beta),y(\beta),z(\beta))$ $f(M)=f(x,y,z)$ $L$ $f(M)$ $L$ 上的第二种曲线积分存在并且

\int_{L_{A B}} f (M) d x = \int_{L +} f (M) d x = \int_{α}^{β} f (x (t), y (t), z (t)) \dot{x} d x .

$y$ $z$ 的积分公式⁵⁰.

$\ointctrclockwise$ $\varointclockwise$ .

2.5 两种曲线积分的关系

\int_{L +} f (M) d x = \int_{0}^{s_{0}} f (x (s), y (s), z (s)) x^{'} (s) d s = \int_{L +} f (M) \cos α_{M} d s .

$y,z$ 同理⁵¹.

2.6 矢量场的环流量, 矢量的曲线积分

$L$ $\vec v(M)=P(M)\vec i+Q(M)\vec j+R(M)\vec k$ $L$ $\vec v$ $L+$ 的曲线积分为

\int_{L +} \vec{v} \cdot d \vec{r} = \int_{L +} (P d x + Q d y + R d z) = \int_{L +} P d x + Q d y + R d z .

前者称为矢量的曲线积分, 后者为矢量场的环流量.

3 曲面积分

3.1 关于面积元素的曲面积分

$Oxyz$ $\Sigma:\left\{\begin{split}x=x(u,v)\\y=y(u,v)\\z=z(u,v)\end{split}\right.,(u,v)\in\Delta$ $\Delta$ $uv$ $f(M)=f(x,y,z)$ $\Sigma$ $f(M)$ $\Sigma$ $\mathrm{d}S=\sqrt{EG-F^2}\mathrm{d}u\mathrm{d}v$ , 且

\iint_{Σ} f (M) d S = \iint_{Δ} f (x (u, v), y (u, v), z (u, v)) \sqrt{E G - F^{2}} d u d v .

3.2 矢量场的通量, 关于面积元素投影的积分

$V$ $\Sigma$ $A$ $B$ $\Sigma$ $M$ $\vec n=\vec n_M$ $B$ $\vec v=\vec v(M)=P(M)\vec i+Q(M)\vec j+R(M)\vec k$ $V$ $\vec v$ $\Sigma$ $A$ $B$ 侧的通量⁵²为

\begin{aligned} \iint_{Σ} \vec{v} \cdot \vec{n} d S & = \iint_{Σ} [P (M) \cos α_{M} + Q (M) \cos β_{M} + R (M) \cos γ_{M}] d S \\ = \iint_{Σ_{B}} P d y d z + Q d z d x + R d x d y . \end{aligned}

这样, 计算矢量场的通量就是要计算上式中间的三个第一种曲面积分, 但由于出现了方向余弦, 这就使得上式右端三个积分有了几何意义. 并且, 由此还可以建立新的曲面积分的概念——第二种曲面积分的概念.

3.3

$\Sigma:\left\{\begin{split}x=x(u,v)\\y=y(u,v)\\z=z(u,v)\end{split}\right.,(u,v)\in\Delta$ $\vec r=\vec r(u,v),(u,v)\in\Delta$ $\Delta$ $uv$ $f(M)=f(x,y,z)$ $\Sigma$ $f(M)$ $\Sigma$ $P,Q,R$ $\Sigma$ 上的连续函数, 则有⁵³公式

\iint_{Σ_{B}} P d y d z + Q d z d x + R d x d y = \iint_{Δ} \pm | \begin{matrix} P & Q & R \\ x_{u}^{'} & y_{u}^{'} & z_{u}^{'} \\ x_{v}^{'} & y_{v}^{'} & z_{v}^{'} \end{matrix} | d u d v .

$\vec r'_u\times\vec r'_v$ $\Sigma$ $B$ 侧而定.

4 Stokes 公式

4.1 Green 公式

$D$ $xOy$ $L$ $P(x,y)$ $Q(x,y)$ $D$ 上有一阶连续偏微商, 则⁵⁴

∳_{L} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y .

$D$ $\ointctrclockwise_Lx\mathrm{d}y=-\ointctrclockwise_Ly\mathrm{d}x=\frac{1}{2}\ointctrclockwise_Lx\mathrm{d}y-y\mathrm{d}x$ .

4.2 Gauss 公式

$V$ $\Sigma$ $P(x,y,z),Q(x,y,z),R(x,y,z)$ $V$ 上有一阶连续偏微商, 则⁵⁵

\underset{Σ_{外}}{∯} P d y d z + Q d z d x + R d x d y = \underset{V}{∭} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) d V .

$V$ $\oiint\limits_{\Sigma_\text{外}}x\mathrm{d}y\mathrm{d}z=\oiint\limits_{\Sigma_\text{外}}y\mathrm{d}z\mathrm{d}x=\oiint\limits_{\Sigma_\text{外}}z\mathrm{d}x\mathrm{d}y=\frac{1}{3}\oiint\limits_{\Sigma_\text{外}}x\mathrm{d}y\mathrm{d}z+y\mathrm{d}z\mathrm{d}x+z\mathrm{d}x\mathrm{d}y$ .

$\vec v$ $M$ $M$ $\Sigma$ $V$ $V$ $\vec n$ $\Sigma$ $\vec v$ $M$ $(\mathrm{div}\,\vec v)_M=\lim\limits_{\Sigma\to M}\frac{\oiint\limits_\Sigma\vec v\cdot\vec n\mathrm{d}S}{V}=(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z})_M$ ⁵⁶ $\mathrm{div}\,\vec v=\nabla\cdot\vec v=\vec i\cdot\frac{\partial\vec v}{\partial x}+\vec j\cdot\frac{\partial\vec v}{\partial y}+\vec k\cdot\frac{\partial\vec v}{\partial z}$ . 此外容易验证:

$\nabla\cdot\varphi\vec v=\varphi\nabla\cdot\vec v+\vec v\cdot\nabla\varphi$

$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}=\nabla^2=\nabla\cdot\nabla$ $\mathrm{div}\,\mathbf{grad}\,u=\Delta u$ .

4.3 Stokes 公式

$\Sigma$ $L$ $P,Q,R$ 有一阶连续偏微商, 则⁵⁷

∳_{L} P d x + Q d y + R d z = \iint_{Σ +} | \begin{matrix} d y d z & d z d x & d x d y \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} | = \iint_{Σ +} | \begin{matrix} \cos α & \cos β & \cos γ \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} | d S .

$\vec v$ $M$ $\mathbf{rot}\,\vec v$ $M$ $\vec n$ $M$ $\vec n$ $L$ $A$ $L$ $\vec t$ $\vec n$ $\vec t$ ⁵⁸ $(\mathbf{rot}\,\vec v)_M\cdot\vec n=\lim\limits_{L\to M}\frac{\oint_L\vec v\cdot\vec t\mathrm{d}s}{A}$ $\vec v=P\vec i+Q\vec j+R\vec k$ $\mathbf{rot}\,\vec v=\left|\begin{array}\\\vec i&\vec j&\vec k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\P&Q&R\end{array}\right|$ ⁵⁹ $\mathbf{rot}\, \vec v=\nabla\times\vec v=\vec i\times\frac{\partial \vec v}{\partial x}+\vec j\times\frac{\partial \vec v}{\partial y}+\vec k\times\frac{\partial \vec v}{\partial z}$ . 此外容易验证:

$\nabla\times\varphi\vec v=\varphi\nabla\times\vec v+\vec v\times\nabla\varphi$
$\nabla\cdot(\vec u\times\vec v)=\vec v\cdot\nabla\times\vec u-\vec u\cdot\nabla\times\vec v$
$\nabla\times(\vec u\times\vec v)=(\vec v\cdot\nabla)\vec u-(\vec u\cdot\nabla)\vec v+\vec u\nabla\cdot\vec v-\vec v\nabla\cdot\vec u$
$\nabla\times(\nabla\times\vec v)=\nabla(\nabla\cdot\vec v)-\Delta\vec v$
$\nabla\times\nabla\vec u\equiv\vec 0$

5 全微分与线积分

5.1 与途径无关的曲线积分

⁶⁰ $\Sigma$ $A,B$ $P,Q$ $\Sigma$ $\int_{(A)}^{(B)}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y$ $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ . 这一条件被称为确切微分条件⁶¹.
空间的曲线积分的确切微分条件与平面情形类似, 只是用 Stokes 公式代替 Green 公式证明就是了.

$(x+y+1)\mathrm{d}x+(x-y^2+3)\mathrm{d}y=0$ ⁶².

6 外微分形式

6.1 外乘积, 外微分形式

$\mathrm{d}x\mathrm{d}y=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\mathrm{d}u\mathrm{d}v$ $\mathrm{d}x\mathrm{d}y=\frac{\partial(x,y)}{\partial(u,v)}\mathrm{d}u\mathrm{d}v$
$y=x$ $\mathrm{d}x\mathrm{d}x=0$
$x,y$ $\mathrm{d}y\mathrm{d}x=-\mathrm{d}x\mathrm{d}y$ .
满足上述两条规则的微分乘积称为微分的外乘积.
由微分的外乘积乘上函数组成的微分形式称为外微分形式, 函数称为微分形式的系数. 有一些性质⁶³.

6.2 Poincaré 引理及其逆

$\omega$ $\mathrm{d}\mathrm{d}\omega=0$ .
$\omega$ $p$ $\mathrm{d}\omega=0$ $p-1$ $\alpha$ $\mathrm{d}\alpha=\omega$ .

6.3

外微分形式的次数	对应的度⁶⁴
0⁶⁵	梯度⁶⁶
1⁶⁷	旋度⁶⁸
2⁶⁹	散度⁷⁰

$\mathrm{d}\mathrm{d}\omega=0$ 有其场论意义.
$\omega$ $\omega=f$ $\mathrm{d}\mathrm{d}f=0$
$\mathbf{rot}\,\mathbf{grad}\,f=\mathbf{0}$ .
$\omega$ $\omega_1=P\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z$ $\mathrm{d}\mathrm{d}\omega_1=0$
$\mathrm{div}\,\mathbf{rot}\,(P,Q,R)=0$ .
$\mathrm{d}\omega=0\Rightarrow\omega=\mathrm{d}\alpha$ 也有其场论意义.
$\vec v$ $\vec v$
$\mathbf{rot}\,\vec v=\mathbf{0}\Leftrightarrow\vec v=\mathbf{grad}\,f$ .
即如果一次微分形式的外微分为零, 则此外微分形式一定是一个函数 (零次外微分形式) 的外微分.
$\vec v$ $\vec v$
$\mathrm{div}\,\vec v=0\Leftrightarrow\vec v=\mathbf{rot}\,\vec b$ .
即如果二次外微分形式的外微分为零, 则此外微分形式一定是一个一次外微分形式的外微分.

6.4

\int_{\partial Σ} ω = \int_{Σ} d ω,

$\omega$ $\Sigma$ $\mathrm{d}\omega$ $\partial\Sigma$ $\Sigma$ $\int$ 表示区域有多少维数就是多少重数.

外微分形式的次数	空间	公式名	公式
0	直线段	Newton - Leibniz 公式	$\omega_0=\int\mathrm{d}\omega_0$
1	平面区域	Green 公式	$\oint\omega_1=\iint\mathrm{d}\omega_1$
1	空间曲面	Stokes 公式	$\oint\omega_1=\iint\mathrm{d}\omega_1$
2	空间中区域	Gauss 公式	$\oiint\omega_2=\iiint\mathrm{d}\omega_2$

最后还要强调的是: 这里所说的 Stokes 公式是微积分的顶峰. 从理论上讲, 这是微积分的终点, 也是微积分从古典走向现代的入口处. 在现代数学中, 微积分的各种定理中, 这条定理也许是用得最多的定理之一. 在数学上, 这是一条少有的简洁, 美丽而深刻的定理.

Part III

To Be Continued …

1 如果

lim_{n \to \infty} b_{n} = B

$\lim\limits_{n\to\infty}b_n = B$ , 那么

lim_{n \to \infty} (a_{n} \pm b_{n}) = A \pm B

$\lim\limits_{n\to\infty}(a_n\pm b_n)=A\pm B$ ,

lim_{n \to \infty} (a_{n} \cdot b_{n}) = A B

$\lim\limits_{n\to\infty}(a_n\cdot b_n)=AB$ ,

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \frac{A}{B}

$\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\frac{A}{B}$

(B \neq 0)

$(B\neq0)$ .
如果

k

$k$ 为常数, 那么

lim_{n \to \infty} k \cdot a_{n} = k A

$\lim\limits_{n\to\infty}{k\cdot a_n}=kA$ .
再如果有

lim_{n \to \infty} c_{n} = C

$\lim\limits_{n\to\infty}c_n=C$ , 且

a_{n} ⩽ b_{n} ⩽ c_{n}

$a_n\leqslant b_n\leqslant c_n$ , 则

A ⩽ B ⩽ C

$A\leqslant B\leqslant C$ . ↩

2 如果

lim_{x \to a} g (x)

$\lim\limits_{x\to a}g(x)$ 也存在且有限, 则有

lim_{x \to a} (f (x) \pm g (x)) = lim_{x \to a} f (x) \pm lim_{x \to a} g (x)

$\lim\limits_{x\to a}(f(x)\pm g(x))=\lim\limits_{x\to a}{f(x)}\pm\lim\limits_{x\to a}g(x)$ ,

lim_{x \to a} (f (x) \cdot g (x)) = lim_{x \to a} f (x) \cdot lim_{x \to a} g (x)

$\lim\limits_{x\to a}(f(x)\cdot g(x))=\lim\limits_{x\to a}{f(x)}\cdot\lim\limits_{x\to a}g(x)$ ,

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} f (x) / lim_{x \to a} g (x)

$\lim\limits_{x\to a}\frac{f(x)}{g(x)}=\lim\limits_{x\to a}{f(x)}/\lim\limits_{x\to a}g(x)$

(lim_{x \to a} g (x) \neq 0)

$(\lim\limits_{x\to a}g(x)\neq0)$ .
此外, 如在点

a

$a$ 的附近成立不等式

f (x) ⩽ g (x) ⩽ h (x)

$f(x)\leqslant g(x)\leqslant h(x)$ 且

lim_{x \to a} f (x) = lim_{x \to a} h (x) = A

$\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}h(x)=A$ , 则

lim_{x \to a} g (x) = A

$\lim\limits_{x\to a}g(x)=A$ . ↩

\ln x_{1} + \ln x_{2} = \int_{1}^{x_{1}} \frac{1}{s} d s + \int_{1}^{x_{2}} \frac{1}{s} d s = \int_{1}^{x_{1}} \frac{1}{s} d s + \int_{x_{1}}^{x_{1} x_{2}} \frac{1}{\frac{s}{x_{1}}} d \frac{s}{x_{1}} = \int_{1}^{x_{1}} \frac{1}{s} d s + \int_{x_{1}}^{x_{1} x_{2}} \frac{1}{s} d s = \int_{1}^{x_{1} x_{2}} \frac{1}{s} d s = \ln x_{1} x_{2}

$\ln x_1+\ln x_2=\int_1^{x_1}\frac{1}{s}{\mathrm{d}s}+\int_1^{x_2}\frac{1}{s}{\mathrm{d}s}=\int_1^{x_1}\frac{1}{s}{\mathrm{d}s}+\int_{x_1}^{x_1 x_2}\frac{1}{\frac{s}{x_1}}{\mathrm{d}{\frac{s}{x_1}}}=\int_1^{x_1}\frac{1}{s}{\mathrm{d}s}+\int_{x_1}^{x_1 x_2}\frac{1}{s}{\mathrm{d}s}=\int_1^{x_1 x_2}\frac{1}{s}{\mathrm{d}s}=\ln {x_1 x_2}$ ↩

f (x + Δ x) - f (x) = \frac{f (x + Δ x) - f (x)}{Δ x} \cdot Δ x

$f(x+\Delta x)-f(x)=\frac{f(x+\Delta x)-f(x)}{\Delta x}\cdot \Delta x$ ,
令

Δ x \to 0

$\Delta x\to0$ , 两边取极限, 由于

lim_{Δ x \to 0} \frac{f (x_{0} + Δ x) - f (x)}{Δ x}

$\lim\limits_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x)}{\Delta x}$ 存在, 所有等式的右边显然趋于零, 故

lim_{Δ x \to 0} [f (x + Δ x) - f (x)] = 0

$\lim\limits_{\Delta x\to0}[f(x+\Delta x)-f(x)]=0$ ,
即

f (x)

$f(x)$ 在

x_{0}

$x_0$ 处连续. ↩

5 由微商的定义,

\frac{1}{x} = \frac{d \ln x}{d x} = \frac{d \log_{e} x}{d x} = lim_{Δ x \to 0} \frac{\log_{e} (x_{0} + Δ x) - \log_{e} x}{Δ x} = lim_{Δ x \to 0} \log_{e} (1 + \frac{Δ x}{x})^{\frac{1}{Δ x}}

$\frac{1}{x}=\frac{\mathrm{d}{\ln x}}{\mathrm{d}x}=\frac{\mathrm{d}{\log_e x}}{\mathrm{d}x}=\lim\limits_{\Delta x\to0}\frac{\log_e(x_0+\Delta x)-\log_e x}{\Delta x}=\lim\limits_{\Delta x\to0}{\log_e{(1+\frac{\Delta x}{x})^\frac{1}{\Delta x}}}$ ,
特别取

x = 1, Δ x = \frac{1}{n}

$x=1,\Delta x=\frac{1}{n}$ , 因此,

lim_{n \to \infty} \log_{e} (1 + \frac{1}{n})^{n} = 1

$\lim\limits_{n\to\infty}{\log_e{(1+\frac{1}{n})^n}}=1$ , 易得 ↩

6 一般来说, 当

y = f (x)

$y=f(x)$ 比较复杂时,

Δ y

$\Delta y$ 不易计算,
然而由微商的定义知道

lim_{Δ x \to 0} \frac{Δ y}{Δ x} = lim_{Δ x \to 0} \frac{f (x_{0} + Δ x) - f (x)}{Δ x} = f^{'} (x_{0})

$\lim\limits_{\Delta x\to0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x)}{\Delta x}=f'(x_0)$ .
若令

\frac{Δ y}{Δ x} - f^{'} (x) = α

$\frac{\Delta y}{\Delta x}-f'(x)=\alpha$ , 则显然有

lim_{Δ x \to 0} α = 0

$\lim\limits_{\Delta x\to0}\alpha=0$ ,
也就是说

Δ y = f^{'} (x) Δ x + α Δ x = d y + α Δ x

$\Delta y=f'(x)\Delta x+\alpha \Delta x=\mathrm{d}y+\alpha \Delta x$ 中的

α Δ x

$\alpha\Delta x$ 可以忽略不计. ↩

7 特别取函数

y = x

$y=x$ , 则

f^{'} (x) = x^{'} = 1

$f'(x)=x'=1$ .
因此, 函数

y = x

$y=x$ 的微分

d x = 1 \cdot Δ x = Δ x

$\mathrm{d}x=1\cdot\Delta x=\Delta x$ . 易得 ↩

8 如果

f (x)

$f(x)$ 在

x_{0}

$x_0$ 点可微, 且在该点取到极值, 那么必有

f^{'} (x_{0}) = 0

$f'(x_0)=0$ . ↩

9 若

f (x)

$f(x)$ 在

[a, b]

$[a,b]$ 上连续, 在

(a, b)

$(a,b)$ 上可微, 且

f (a) = f (b)

$f(a)=f(b)$ , 则至少存在一点

ξ \in (a, b)

$\xi\in(a,b)$ , 使得

f^{'} (ξ) = 0

$f'(\xi)=0$ . ↩

10 设函数

f (x)

$f(x)$ 和

g (x)

$g(x)$ 都是

[a, b]

$[a,b]$ 上的连续函数, 且在

(a, b)

$(a,b)$ 上可微, 且

\forall x \in (a, b), g^{'} (x) \neq 0

$\forall x\in(a,b),g'(x)\neq0$ , 则必有

ξ \in (a, b)

$\xi\in(a,b)$ , 使得

\frac{f (b) - f (a)}{g (b) - g (a)} = \frac{f^{'} (ξ)}{g^{'} (ξ)}

$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}$ . ↩

11 由于

\frac{Δ y}{Δ x} = \frac{Δ y}{Δ u} \cdot \frac{Δ u}{Δ x}

$\frac{\Delta y}{\Delta x}=\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x}$ ,
令

Δ x \to 0

$\Delta x\to0$ , 注意这时也有

Δ u \to 0

$\Delta u\to0$ , 于是得

y^{'} = lim_{Δ x \to 0} \frac{Δ y}{Δ x} = lim_{Δ x \to 0} (\frac{Δ y}{Δ u} \cdot \frac{Δ u}{Δ x}) = lim_{Δ x \to 0} \frac{Δ y}{Δ u} \cdot lim_{Δ x \to 0} \frac{Δ u}{Δ x} = lim_{Δ u \to 0} \frac{Δ y}{Δ u} \cdot lim_{Δ x \to 0} \frac{Δ u}{Δ x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

$y'=\lim\limits_{\Delta x\to0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\to0}(\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x})=\lim\limits_{\Delta x\to0}\frac{\Delta y}{\Delta u}\cdot\lim\limits_{\Delta x\to0}\frac{\Delta u}{\Delta x}=\lim\limits_{\Delta u\to0}\frac{\Delta y}{\Delta u}\cdot\lim\limits_{\Delta x\to0}\frac{\Delta u}{\Delta x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}$ . ↩

12 由于

y = f (g (y))

$y=f(g(y))$ , 立刻有

1 = f^{'} (x) g^{'} (y)

$1=f'(x)g'(y)$ . ↩

(\arcsin x)^{'} = \frac{1}{(\sin y)^{'}} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^{2} y}} = \frac{1}{\sqrt{1 - x^{2}}}

$(\arcsin x)'=\frac{1}{(\sin y)'}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-\sin^2y}}=\frac{1}{\sqrt{1-x^2}}$ . ↩

(u^{v})^{'} = u^{v} [v^{'} \ln u + v \frac{u^{'}}{u}]

$(u^v)'=u^v[v'\ln u+v\frac{u'}{u}]$

(\ln (1 + x))^{(n)} = (- 1)^{n - 1} \frac{(n - 1)!}{(1 + x)^{n}}

$(\ln (1+x))^{(n)}=(-1)^{n-1}\frac{(n-1)!}{(1+x)^n}$

(\sin x)^{(n)} = \sin (x + \frac{n π}{2})

$(\sin x)^{(n)}=\sin (x+\frac{n\pi}{2})$

(\cos x)^{(n)} = \cos (x + \frac{n π}{2})

$(\cos x)^{(n)}=\cos (x+\frac{n\pi}{2})$ ↩

15 设

y = f (x), x = φ (t)

$y=f(x),x=\varphi(t)$ 都是可微函数,
而已知复合函数

y = f [φ (t)]

$y=f[\varphi(t)]$ 的微商为

y^{'} = f^{'} (x) φ^{'} (t)

$y'=f'(x)\varphi'(t)$ ,
因此

d y = f^{'} (x) φ^{'} (t) d t

$\mathrm{d}y=f'(x)\varphi'(t)\mathrm{d}t$ ,
但

d x = φ^{'} (t) d t

$\mathrm{d}x=\varphi'(t)\mathrm{d}t$ ,
故得

d y = f^{'} (x) d x

$\mathrm{d}y=f'(x)\mathrm{d}x$ .
这就是说, 不论

x

$x$ 是自变量或中间变量,

f (x)

$f(x)$ 的微分形式

d y = f^{'} (x) d x

$\mathrm{d}y=f'(x)\mathrm{d}x$ 总是不变的. ↩

16 以二阶微分为例, 命

y = f (x)

$y=f(x)$ , 则

d y = f^{'} (x) d x

$\mathrm{d}y=f'(x)\mathrm{d}x$ ,
则有

d^{2} y = d (d y) = d [f^{'} (x) d x] = f^{″} (x) d x^{2}

$\mathrm{d}^2y=\mathrm{d}(\mathrm{d}y)=\mathrm{d}[f'(x)\mathrm{d}x]=f''(x)\mathrm{d}x^2$ . ↩

(u v)^{(n)} = \sum_{r = 0}^{n} C_{n}^{r} u^{(n - r)} v^{(r)}

$(uv)^{(n)}=\sum_{r=0}^n{\mathrm{C}_n^ru^{(n-r)}v^{(r)}}$ ↩

\frac{d y}{d x} = \frac{d y}{d t} \frac{d t}{d x} = ψ^{'} (t) \frac{1}{φ^{'} (t)} = \frac{ψ^{'} (t)}{φ^{'} (t)}

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}x}=\psi'(t)\frac{1}{\varphi'(t)}=\frac{\psi'(t)}{\varphi'(t)}$ ,

\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{ψ^{'} (t)}{φ^{'} (t)}) = \frac{d}{d t} (\frac{ψ^{'} (t)}{φ^{'} (t)}) \frac{d t}{d x} = \frac{ψ^{″} (t) φ^{'} (t) - ψ^{'} (t) φ^{″} (t)}{[φ^{'} (t)]^{2}} \frac{1}{φ^{'} (t)} = \frac{ψ^{″} φ^{'} - ψ^{'} φ^{″}}{φ^{' 3}}

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\psi'(t)}{\varphi'(t)})=\frac{\mathrm{d}}{\mathrm{d}t}(\frac{\psi'(t)}{\varphi'(t)})\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{\psi''(t)\varphi'(t)-\psi'(t)\varphi''(t)}{[\varphi'(t)]^2}\frac{1}{\varphi'(t)}=\frac{\psi''\varphi'-\psi'\varphi''}{\varphi'^3}$ . ↩

19 令

P_{n} (x) = f (0) + f^{'} (0) x + \frac{f^{″} (0)}{2!} x^{2} + \dots + \frac{f^{(n)} (0)}{n!} x^{n}

$P_n(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n$ ,
记

f (x) - P_{n} (x) = \frac{x^{n + 1}}{(n + 1)!} Q_{n}

$f(x)-P_n(x)=\frac{x^{n+1}}{(n+1)!}Q_n$ ,
作

z

$z$ 的函数

ϕ (z) = f (x) - [f (z) + (x - z) f^{'} (z) + \frac{(x - z)^{2}}{2!} f^{″} (z) + \dots + \frac{(x - z)^{n}}{n!} f^{(n)} (z) + \frac{(x - z)^{n + 1}}{(n + 1)!} Q_{n}]

$\phi(z)=f(x)-[f(z)+(x-z)f'(z)+\frac{(x-z)^2}{2!}f''(z)+\cdots+\frac{(x-z)^n}{n!}f^{(n)}(z)+\frac{(x-z)^{n+1}}{(n+1)!}Q_n]$ ,
则显然有

ϕ (0) = ϕ (x) = 0

$\phi(0)=\phi(x)=0$ . 由微分中值定理, 一定有点

ξ = θ x, 0 < θ < 1

$\xi=\theta x,0<\theta<1$ , 使

ϕ^{'} (ξ) = 0

$\phi'(\xi)=0$ ,
但

ϕ^{'} (z) = \frac{(x - z)^{n}}{n!} [Q_{n} - f^{(n + 1)} (z)]

$\phi'(z)=\frac{(x-z)^n}{n!}[Q_n-f^{(n+1)}(z)]$ , 故

Q_{n} = f^{(n + 1)} (ξ)

$Q_n=f^{(n+1)}(\xi)$ .

Q_{n}

$Q_n$ 称为泰勒展开式的余项, 所以

f (x) - P_{n} = O (x^{n + 1})

$f(x)-P_n=O(x^{n+1})$ . ↩

20 若

f (x), g (x)

$f(x),g(x)$ 在

x = x_{0}

$x=x_0$ 处有二阶微商,

g^{'} (x_{0}) \neq 0

$g'(x_0)\neq0$ , 且

lim_{x \to x_{0}} f (x) = 0, lim_{x \to x_{0}} g (x) = 0

$\lim\limits_{x\to x_0}f(x)=0,\lim\limits_{x\to x_0}g(x)=0$ ,
则

lim_{x \to x_{0}} \frac{f (x)}{g (x)} = lim_{x \to x_{0}} \frac{f^{'} (x_{0}) (x - x_{0}) + O [(x - x_{0})^{2}]}{g^{'} (x_{0}) (x - x_{0}) + O [(x - x_{0})^{2}]} = lim_{x \to x_{0}} \frac{f^{'} (x_{0}) + O (x - x_{0})}{g^{'} (x_{0}) + O (x - x_{0})} = lim_{x \to x_{0}} \frac{f^{'} (x_{0})}{g^{'} (x_{0})}

$\lim\limits_{x\to x_0}\frac{f(x)}{g(x)}=\lim\limits_{x\to x_0}\frac{f'(x_0)(x-x_0)+O[(x-x_0)^2]}{g'(x_0)(x-x_0)+O[(x-x_0)^2]}=\lim\limits_{x\to x_0}\frac{f'(x_0)+O(x-x_0)}{g'(x_0)+O(x-x_0)}=\lim\limits_{x\to x_0}\frac{f'(x_0)}{g'(x_0)}$ .
当然还可以有: 若

f (x), g (x)

$f(x),g(x)$ 在

x = x_{0}

$x=x_0$ 处有

n

$n$ 阶微商, 且

lim_{x \to x_{0}} f (x) = lim_{x \to x_{0}} f^{'} (x) = \dots = lim_{x \to x_{0}} f^{(n - 1)} (x) = 0,

$\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0}f'(x)=\cdots=\lim\limits_{x\to x_0}f^{(n-1)}(x)=0,$

lim_{x \to x_{0}} g (x) = lim_{x \to x_{0}} g^{'} (x) = \dots = lim_{x \to x_{0}} g^{(n - 1)} (x) = 0

$\lim\limits_{x\to x_0}g(x)=\lim\limits_{x\to x_0}g'(x)=\cdots=\lim\limits_{x\to x_0}g^{(n-1)}(x)=0$ ,
则

lim_{x \to x_{0}} \frac{f (x)}{g (x)} = \frac{f^{(n)} (x_{0})}{g^{(n)} (x_{0})}

$\lim\limits_{x\to x_0}\frac{f(x)}{g(x)}=\frac{f^{(n)}(x_0)}{g^{(n)}(x_0)}$ . ↩

a_{0} (x) \frac{d^{n} y}{d x^{n}} + a_{1} (x) \frac{d^{n - 1} y}{d x^{n - 1}} + \dots + a_{n} (x) y = f (x)

$a_0(x)\frac{\mathrm{d}^ny}{\mathrm{d}x^n}+a_1(x)\frac{\mathrm{d}^{n-1}y}{\mathrm{d}x^{n-1}}+\cdots+a_n(x)y=f(x)$ ↩

g (y) d y = f (x) d x \Rightarrow \int g (y) d y = \int f (x) d x

$g(y)\mathrm{d}y=f(x)\mathrm{d}x\Rightarrow\int g(y)\mathrm{d}y=\int f(x)\mathrm{d}x$ ↩

23 令

y = x u

$y=xu$ 引入新变量,
代回

y^{'} = f (\frac{y}{x})

$y'=f(\frac{y}{x})$ 得到

u + x u^{'} = f (u)

$u+xu'=f(u)$ ,
用分离变量的方法解得

x = e^{\int \frac{d u}{f (u) - u}}

$x=e^{\int\frac{\mathrm{d}u}{f(u)-u}}$ . ↩

24 考虑方程 (1)

d (ψ (x) y) = f (x) d x

$\mathrm{d}(\psi(x)y)=f(x)\mathrm{d}x$ ,
将上式两边积分, 得

ψ (x) y = \int f (x) d x \Rightarrow y = \frac{1}{ψ (x)} \int f (x) d x

$\psi(x)y=\int f(x)\mathrm{d}x\Rightarrow y=\frac{1}{\psi(x)}\int f(x)\mathrm{d}x$ ,
原方程 (1) 就是

ψ (x) \frac{d y}{d x} + ψ^{'} (x) y = f (x)

$\psi(x)\frac{\mathrm{d}y}{\mathrm{d}x}+\psi'(x)y=f(x)$ ,
方程 (1) 还可以写成

\frac{d y}{d x} + P (x) y = Q (x)

$\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)$ 的形式, 此处

P (x) = \frac{ψ^{'} (x)}{ψ (x)}, Q (x) = \frac{f (x)}{ψ (x)}

$P(x)=\frac{\psi'(x)}{\psi(x)},Q(x)=\frac{f(x)}{\psi(x)}$ .
显然

ψ (x)

$\psi(x)$ 可以轻松解出, 进而算出

f (x)

$f(x)$ , 代入方程 (1) 的解即可. ↩

\frac{d y}{d x} + P (x) y = 0 \Rightarrow y = C e^{- \int P (x) d x}

$\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=0\Rightarrow y=Ce^{-\int P(x)\mathrm{d}x}$ , 这里

\int P (x) d x

$\int P(x)\mathrm{d}x$ 仅表示

P (x)

$P(x)$ 的一个原函数.
将常数

C

$C$ 变易为

x

$x$ 的函数

C (x)

$C(x)$ , 然后将

C (x) e^{- \int P (x) d x}

$C(x)e^{-\int P(x)\mathrm{d}x}$ 代入

\frac{d y}{d x} + P (x) y = Q (x)

$\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)$ ,
就得到一个

C (x)

$C(x)$ 的微分方程

C^{'} (x) = Q (x) e^{\int P (x) d x}

$C'(x)=Q(x)e^{\int P(x)\mathrm{d}x}$ ,

C (x) = \int Q (x) e^{\int P (x) d x} d x

$C(x)=\int Q(x)e^{\int P(x)\mathrm{d}x}\mathrm{d}x$ 就是它的解.
回代即可. ↩

26 令

p = y^{'}

$p=y'$ 即可 ↩

p = y^{'} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d p}{d x} = \frac{d p}{d y} \cdot \frac{d y}{d x} = p \frac{d p}{d y}

$p=y'\Rightarrow\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{\mathrm{d}p}{\mathrm{d}y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}=p\frac{\mathrm{d}p}{\mathrm{d}y}$ 可将方程化为

\frac{d p}{d y} = φ (y, p)

$\frac{\mathrm{d}p}{\mathrm{d}y}=\varphi(y,p)$ 的形式 ↩

28 线性相关意即, 要存在两个不同时为零的常数

C_{1}, C_{2}

$C_1,C_2$ ,
使得

C_{1} φ (x) + C_{2} ψ (x) \equiv 0

$C_1\varphi(x)+C_2\psi(x)\equiv0$ ,
顺便有

C_{1} φ^{'} (x) + C_{2} ψ^{'} (x) \equiv 0

$C_1\varphi'(x)+C_2\psi'(x)\equiv0$ ,
当且仅当

| \begin{matrix} φ (x) & ψ (x) \\ φ^{'} (x) & ψ^{'} (x) \end{matrix} | = 0

$\left|\begin{array}\\\varphi(x)&\psi(x)\\\varphi'(x)&\psi'(x)\end{array}\right|=0$ . ↩

29 例如

φ (x) = {\begin{matrix} - x & x ⩽ 0 \\ 0 & x ⩾ 0 \end{matrix},

$\varphi(x)=\left\{\begin{array}\\-x&x\leqslant0\\0&x\geqslant0\end{array}\right.,$

ψ (x) = {\begin{matrix} 0 & x ⩽ 0 \\ x & x ⩾ 0 \end{matrix} .

$\psi(x)=\left\{\begin{array}\\0&x\leqslant0\\x&x\geqslant0\end{array}\right..$ ↩

W (φ (x_{0}), ψ (x_{0})) = 0 \Rightarrow {\begin{matrix} C_{1} φ (x_{0}) + C_{2} ψ (x_{0}) = 0 \\ C_{1} φ^{'} (x_{0}) + C_{2} ψ^{'} (x_{0}) = 0 \end{matrix} .

$W(\varphi(x_0),\psi(x_0))=0\Rightarrow\left\{\begin{array}\\C_1\varphi(x_0)+C_2\psi(x_0)=0\\C_1\varphi'(x_0)+C_2\psi'(x_0)=0\end{array}\right..$
考虑函数

y (x) = C_{1} φ (x) + C_{2} ψ (x)

$y(x)=C_1\varphi(x)+C_2\psi(x)$ , 它也是齐次方程的解, 且符合初始条件

y (x_{0}) = 0, y^{'} (x_{0}) = 0

$y(x_0)=0,y'(x_0)=0$ ;
另一方面,

\tilde{y} (x) \equiv 0

$\tilde{y}(x)\equiv0$ 当然是齐次方程的解, 和

y (x)

$y(x)$ 满足同样的初始条件.
由解的存在唯一性定理推知

y (x) \equiv \tilde{y} (x) \equiv 0

$y(x)\equiv\tilde{y}(x)\equiv0$ , 这就证明了

φ (x), ψ (x)

$\varphi(x),\psi(x)$ 线性相关. ↩

\frac{d W}{d x} = \frac{d}{d x} (y_{1} y_{2}^{'} - y_{1}^{'} y_{2}) = y_{1} y_{2}^{″} - y_{1}^{″} y_{2}

$\frac{\mathrm{d}W}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(y_1y_2'-y_1'y_2)=y_1y_2''-y_1''y_2$ ,
但

y_{1}^{″} = - p (x) y_{1}^{'} - q (x) y_{1}, y_{2}^{″} = - p (x) y_{2}^{'} - q (x) y_{2}

$y_1''=-p(x)y_1'-q(x)y_1,y_2''=-p(x)y_2'-q(x)y_2$ ,
所以

\frac{d W}{d x} = y_{1} (- p y_{2}^{'} - q y_{2}) - y_{2} (- p y_{1}^{'} - q y_{1}) = - p W

$\frac{\mathrm{d}W}{\mathrm{d}x}=y_1(-py_2'-qy_2)-y_2(-py_1'-qy_1)=-pW$ . ↩

W (y_{1}, y_{2}) = e^{- \int P d x}

$W(y_1,y_2)=e^{-\int P\mathrm{d}x}$ 的两边同除以

y_{1}^{2}

$y_1^2$ ,
得

\frac{y_{1} y_{2}^{'} - y_{1}^{'} y_{2}}{y_{1}^{2}} = \frac{1}{y_{1}^{2}} e^{- \int P d x} = \frac{d}{d x} (\frac{y_{2}}{y_{1}})

$\frac{y_1y_2'-y_1'y_2}{y_1^2}=\frac{1}{y_1^2}e^{-\int P\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\frac{y_2}{y_1})$ ,
积分之, 得

\frac{y_{2}}{y_{1}} = \int \frac{1}{y_{1}^{2}} e^{- \int P d x} d x

$\frac{y_2}{y_1}=\int\frac{1}{y_1^2}e^{-\int P\mathrm{d}x}\mathrm{d}x$ ↩

33 假设有形如

y^{*} = C_{1} (x) y_{1} (x) + C_{2} (x) y_{2} (x)

$y^\ast=C_1(x)y_1(x)+C_2(x)y_2(x)$ 的特解.
于是

\frac{d y^{*}}{d x} = C_{1} y_{1}^{'} + C_{2} y_{2}^{'} + [C_{1}^{'} y_{1} + C_{2}^{'} y_{2}]

$\frac{\mathrm{d}y^\ast}{\mathrm{d}x}=C_1y_1'+C_2y_2'+[C_1'y_1+C_2'y_2]$ , 不妨命

C_{1}^{'} y_{1} + C_{2}^{'} y_{2} = 0

$C_1'y_1+C_2'y_2=0$ .
于是

\frac{d^{2} y^{*}}{d x^{2}} = C_{1} y_{1}^{″} + C_{2} y_{2}^{″} + [C_{1}^{'} y_{1}^{'} + C_{2}^{'} y_{2}^{'}]

$\frac{\mathrm{d}^2y^\ast}{\mathrm{d}x^2}=C_1y_1''+C_2y_2''+[C_1'y_1'+C_2'y_2']$ , 代入

y^{″} + p y^{'} + q y = f

$y''+py'+qy=f$ ,
得到

C_{1} [y_{1}^{″} + p y_{1}^{'} + q y_{1}] + C_{2} [y_{2}^{″} + p y_{2}^{'} + q y_{2}] + [C_{1}^{'} y_{1}^{'} + C_{2}^{'} y_{2}^{'}] = f

$C_1[y_1''+py_1'+qy_1]+C_2[y_2''+py_2'+qy_2]+[C_1'y_1'+C_2'y_2']=f$ .
由于

y_{1}, y_{2}

$y_1,y_2$ 是齐次方程的解, 所以

C_{1}^{'} y_{1}^{'} + C_{2}^{'} y_{2}^{'} = f

$C_1'y_1'+C_2'y_2'=f$ .
将前式与

C_{1}^{'} y_{1} + C_{2}^{'} y_{2} = 0

$C_1'y_1+C_2'y_2=0$ 联立, 由

W (y_{1}, y_{2}) \neq 0

$W(y_1,y_2)\neq0$ 可知有解,
解方程得

{\begin{matrix} C_{1}^{'} (x) = \frac{- y_{2} (x) f (x)}{y_{1} (x) y_{2}^{'} (x) - y_{1}^{'} (x) y_{2} (x)} \\ C_{2}^{'} (x) = \frac{y_{1} (x) f (x)}{y_{1} (x) y_{2}^{'} (x) - y_{1}^{'} (x) y_{2} (x)} \end{matrix}

$\left\{\begin{array}\\C_1'(x)=\frac{-y_2(x)f(x)}{y_1(x)y_2'(x)-y_1'(x)y_2(x)}\\C_2'(x)=\frac{y_1(x)f(x)}{y_1(x)y_2'(x)-y_1'(x)y_2(x)}\end{array}\right.$ . 积分之即可. ↩

34 前面说过, 已知齐次二阶线性常微分方程的一个非平凡解时, 可以算出另一个线性无关的解.
故

y_{2} (x) = e^{- \frac{a}{2} x} \int e^{a x} \cdot e^{- \int a d x} d x

$y_2(x)=e^{-\frac{a}{2}x}\int{e^{ax}\cdot e^{-\int a\mathrm{d}x}\mathrm{d}x}$ . ↩

35 由

λ

$\lambda$ 得出复值解

\tilde{y} (x) = e^{(α \pm i β) x}

$\tilde{y}(x)=e^{(\alpha\pm\mathrm{i}\beta)x}$ , 则应用 Euler 公式即可.

y_{1} = \frac{1}{2} (e^{(α + i β) x} + e^{(α - i β) x}) = e^{a x} \cos β x,

$y_1=\frac{1}{2}(e^{(\alpha+\mathrm{i}\beta)x}+e^{(\alpha-\mathrm{i}\beta)x})=e^{ax}\cos\beta x,$

y_{2} = \frac{1}{2 i} (e^{(α + i β) x} - e^{(α - i β) x}) = e^{a x} \sin β x .

$y_2=\frac{1}{2\mathrm{i}}(e^{(\alpha+\mathrm{i}\beta)x}-e^{(\alpha-\mathrm{i}\beta)x})=e^{ax}\sin\beta x.$ ↩

d (u \pm v) = d u \pm d v

$\mathrm{d}(u\pm v)=\mathrm{d}u\pm\mathrm{d}v$ ,

d (u v) = u d v + v d u

$\mathrm{d}(uv)=u\mathrm{d}v+v\mathrm{d}u$ ,

d (\frac{u}{v}) = \frac{v d u - u d v}{v^{2}}

$\mathrm{d}(\frac{u}{v})=\frac{v\mathrm{d}u-u\mathrm{d}v}{v^2}$ . ↩

37 如果

x, y

$x,y$ 是

u, v

$u,v$ 的函数, 则

\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}

$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}$ ,

\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial v}

$\frac{\partial f}{\partial v}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}$ .
计算出对

u, v

$u,v$ 的全微分

d x

$\mathrm{d}x$ 与

d y

$\mathrm{d}y$ ,
立刻推出

d f = \frac{\partial f}{\partial u} d u + \frac{\partial f}{\partial v} d v = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y

$\mathrm{d}f=\frac{\partial f}{\partial u}\mathrm{d}u+\frac{\partial f}{\partial v}\mathrm{d}v=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y$ . ↩

38 主要是因为

Δ_{x} Δ_{y} f = Δ_{y} Δ_{x} f

$\Delta_x\Delta_yf=\Delta_y\Delta_xf$ . ↩

F (x, y, z (x, y)) \equiv 0 \Rightarrow F_{x}^{'} + F_{z}^{'} \frac{\partial z}{\partial x} \equiv 0

$F(x,y,z(x,y))\equiv0\Rightarrow F'_x+F'_z\frac{\partial z}{\partial x}\equiv0$ ↩

{\begin{matrix} F (x, y, \dots, u (x, y, \dots), v (x, y, \dots)) = 0 \\ G (x, y, \dots, u (x, y, \dots), v (x, y, \dots)) = 0 \end{matrix} \Rightarrow {\begin{matrix} \frac{\partial F}{\partial x} + \frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x} = 0 \\ \frac{\partial G}{\partial x} + \frac{\partial G}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial G}{\partial v} \frac{\partial v}{\partial x} = 0 \end{matrix}

$\left\{\begin{array}\\F(x,y,\cdots,u(x,y,\cdots),v(x,y,\cdots))=0\\G(x,y,\cdots,u(x,y,\cdots),v(x,y,\cdots))=0\end{array}\right.\Rightarrow\left\{\begin{array}\\\frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x}=0\\\frac{\partial G}{\partial x}+\frac{\partial G}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial x}=0\end{array}\right.$ ↩

ρ d ρ d θ

$\rho\mathrm{d}\rho\mathrm{d}\theta$ ↩

{\begin{aligned} x = ρ \sin θ \cos φ \\ y = ρ \sin θ \sin φ \\ z = ρ \cos θ \end{aligned}

$\left\{\begin{split}&x=\rho\sin\theta\cos\varphi\\&y=\rho\sin\theta\sin\varphi\\&z=\rho\cos\theta\end{split}\right.$ ,

φ

$\varphi$ 为

\vec{ρ}

$\vec\rho$ 在

x O y

$xOy$ 上的投影与

O x

$Ox$ 的夹角,

θ

$\theta$ 为

\vec{ρ}

$\vec\rho$ 与

O z

$Oz$ 的夹角.
体素由

\frac{\partial (x, y, z)}{\partial (ρ, θ, φ)} = ρ^{2} \sin θ

$\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}=\rho^2\sin\theta$ 求得. ↩

{\begin{aligned} x = r \cos θ \\ y = r \sin θ \\ z = z \end{aligned}

$\left\{\begin{split}&x=r\cos\theta\\&y=r\sin\theta\\&z=z\end{split}\right.$ 的体素为

r d r d θ d z

$r\mathrm{d}r\mathrm{d}\theta\mathrm{d}z$ ↩

{\begin{array}{r} x, y, z > 0 \\ x + y + z < a \end{array}

$\left\{\begin{split}x,y,z>0\\x+y+z<a\end{split}\right.$ 确定了坐标面与

x + y + z = a

$x+y+z=a$ 之间的四面体,
引入使得

{\begin{array}{r} x + y + z = q_{1} \\ a (y + z) = q_{1} q_{2} \\ a^{2} z = q_{1} q_{2} q_{3} \end{array}

$\left\{\begin{split}x+y+z=q_1\\a(y+z)=q_1q_2\\a^2z=q_1q_2q_3\end{split}\right.$ 成立的新变量

{\begin{aligned} q_{1} = x + y + z \\ q_{2} = \frac{a (y + z)}{x + y + z} \\ q_{3} = \frac{a z}{y + z} \end{aligned}

$\left\{\begin{split}&q_1=x+y+z\\&q_2=\frac{a(y+z)}{x+y+z}\\&q_3=\frac{az}{y+z}\end{split}\right.$ ,
容易算出

{\begin{aligned} x = \frac{q_{1} (a - q_{2})}{a} \\ y = \frac{q_{1} q_{2} (a - q_{3})}{a^{2}} \\ z = \frac{q_{1} q_{2} q_{3}}{a^{2}} \end{aligned}

$\left\{\begin{split}&x=\frac{q_1(a-q_2)}{a}\\&y=\frac{q_1q_2(a-q_3)}{a^2}\\&z=\frac{q_1q_2q_3}{a^2}\end{split}\right.$ 且

0 < q_{1}, q_{2}, q_{3} < a

$0<q_1,q_2,q_3<a$ 以及

\frac{\partial (x, y, z)}{\partial (q_{1}, q_{2}, q_{3})} = \frac{1}{a^{3}} q_{1}^{2} q_{2}

$\frac{\partial(x,y,z)}{\partial(q_1,q_2,q_3)}=\frac{1}{a^3}q_1^2q_2$ .
则

\int_{0}^{a} d x \int_{0}^{a - x} d y \int_{0}^{a - x - y} f (x, y, z) d z = \frac{1}{a^{3}} \int_{0}^{a} q_{1}^{2} d q_{1} \int_{0}^{a} q_{2} d q_{2} \int_{0}^{a} f (x (q_{1}, q_{2}, q_{3}), y (q_{1}, q_{2}, q_{3}), z (q_{1}, q_{2}, q_{3})) d q_{3}

$\int_0^a\mathrm{d}x\int_0^{a-x}\mathrm{d}y\int_0^{a-x-y}f(x,y,z)\mathrm{d}z=\frac{1}{a^3}\int_0^aq_1^2\mathrm{d}q_1\int_0^aq_2\mathrm{d}q_2\int_0^af(x(q_1,q_2,q_3),y(q_1,q_2,q_3),z(q_1,q_2,q_3))\mathrm{d}q_3$ . ↩

45 易知

d S = | {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} | d u d v

$\mathrm{d}S=|\vec r'_u\times\vec r'_v|\mathrm{d}u\mathrm{d}v$ ,
而

| {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} |^{2} = {\vec{r}}_{u}^{' 2} {\vec{r}}_{v}^{' 2} - ({\vec{r}}_{u}^{'} \cdot {\vec{r}}_{v}^{'})^{2}

$|\vec r'_u\times\vec r'_v|^2=\vec r'^2_u\vec r'^2_v-(\vec r'_u\cdot\vec r'_v)^2$ ,
故令

{\begin{aligned} E = {\vec{r}}_{u}^{' 2} = x_{u}^{' 2} + y_{u}^{' 2} + z_{u}^{' 2} \\ G = {\vec{r}}_{v}^{' 2} = x_{v}^{' 2} + y_{v}^{' 2} + z_{v}^{' 2} \\ F = {\vec{r}}_{u}^{'} \cdot {\vec{r}}_{v}^{'} = x_{u}^{'} x_{v}^{'} + y_{u}^{'} y_{v}^{'} + z_{u}^{'} z_{v}^{'} \end{aligned}

$\left\{\begin{split}&E=\vec r'^2_u=x'^2_u+y'^2_u+z'^2_u\\&G=\vec r'^2_v=x'^2_v+y'^2_v+z'^2_v\\&F=\vec r'_u\cdot\vec r'_v=x'_ux'_v+y'_uy'_v+z'_uz'_v\end{split}\right.$ . ↩

d S = \sqrt{1 + z_{x}^{' 2} + z_{y}^{' 2}} d x d y

$\mathrm{d}S=\sqrt{1+z'^2_x+z'^2_y}\mathrm{d}x\mathrm{d}y$ ↩

47 命

l

$l$ 的方向余弦为

(\cos α, \cos β, \cos γ)

$(\cos\alpha,\cos\beta,\cos\gamma)$ , 则

l

$l$ 上的

M

$M$ 可表示为

(x_{0} + t \cos α, y_{0} + t \cos β, z_{0} + t \cos γ)

$(x_0+t\cos\alpha,y_0+t\cos\beta,z_0+t\cos\gamma)$ .
于是

(\frac{\partial u}{\partial l})_{M_{0}} = lim_{t \to 0} \frac{u (x_{0} + t \cos α, y_{0} + t \cos β, z_{0} + t \cos γ) - u (x_{0}, y_{0}, z_{0})}{t} = \frac{\partial u}{\partial x} \cos α + \frac{\partial u}{\partial y} \cos β + \frac{\partial u}{\partial z} \cos γ

$(\frac{\partial u}{\partial l})_{M_0}=\lim\limits_{t\to0}\frac{u(x_0+t\cos\alpha,y_0+t\cos\beta,z_0+t\cos\gamma)-u(x_0,y_0,z_0)}{t}=\frac{\partial u}{\partial x}\cos\alpha+\frac{\partial u}{\partial y}\cos\beta+\frac{\partial u}{\partial z}\cos\gamma$ ,
则

(\frac{\partial u}{\partial l})_{M_{0}} = (\frac{\partial u}{\partial x} \vec{i} + \frac{\partial u}{\partial y} \vec{j} + \frac{\partial u}{\partial z} \vec{k})_{M_{0}} \cdot (\cos α, \cos β, \cos γ)

$(\frac{\partial u}{\partial l})_{M_0}=(\frac{\partial u}{\partial x}\vec i+\frac{\partial u}{\partial y}\vec j+\frac{\partial u}{\partial z}\vec k)_{M_0}\cdot(\cos\alpha,\cos\beta,\cos\gamma)$ . ↩

48 设流线

\vec{r} (t) = x (t) \vec{i} + y (t) \vec{j} + z (t) \vec{k}

$\vec r(t)=x(t)\vec i+y(t)\vec j+z(t)\vec k$ ,
则

\frac{d \vec{r}}{d t} / / \vec{v} \Rightarrow \frac{d x}{P} = \frac{d y}{Q} = \frac{d z}{R}

$\frac{\mathrm{d}\vec r}{\mathrm{d}t}/\mkern-4mu/\vec v\Rightarrow\frac{\mathrm{d}x}{P}=\frac{\mathrm{d}y}{Q}=\frac{\mathrm{d}z}{R}$ .
这个微分方程组的存在性与唯一性的证明略. ↩

49 在

x O y

$xOy$ 中形如

\int_{L_{A B}} f (M) d y = \int_{L_{A B}} f (x, y) d y

$\int_{L_{AB}}f(M)\mathrm{d}y=\int_{L_{AB}}f(x,y)\mathrm{d}y$ ;
在

O x y z

$Oxyz$ 中形如

\int_{L_{A B}} f (M) d z = \int_{L_{A B}} f (x, y, z) d z

$\int_{L_{AB}}f(M)\mathrm{d}z=\int_{L_{AB}}f(x,y,z)\mathrm{d}z$ . ↩

50 习惯上把参数增加时曲线运行的方向叫做参数曲线的正向. ↩

\int_{L +} f (M) d y = \int_{0}^{s_{0}} f (x (s), y (s), z (s)) y^{'} (s) d s = \int_{L +} f (M) \cos β_{M} d s

$\int_{L+}f(M)\mathrm{d}y=\int_0^{s_0}f(x(s),y(s),z(s))y'(s)\mathrm{d}s=\int_{L+}f(M)\cos\beta_M\mathrm{d}s$ ,

\int_{L +} f (M) d z = \int_{0}^{s_{0}} f (x (s), y (s), z (s)) z^{'} (s) d s = \int_{L +} f (M) \cos γ_{M} d s

$\int_{L+}f(M)\mathrm{d}z=\int_0^{s_0}f(x(s),y(s),z(s))z'(s)\mathrm{d}s=\int_{L+}f(M)\cos\gamma_M\mathrm{d}s$ . ↩

52 或简单叫做在

Σ

$\Sigma$ 的

B

$B$ 侧的通量 ↩

{\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'}

$\vec r'_u\times\vec r'_v$ 是曲面

Σ

$\Sigma$ 的法矢量, 且

\vec{n} = \pm \frac{{\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'}}{| {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} |} = \pm \frac{1}{| {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} |} | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ x_{u}^{'} & y_{u}^{'} & z_{u}^{'} \\ x_{v}^{'} & y_{v}^{'} & z_{v}^{'} \end{matrix} |

$\vec n=\pm\frac{\vec r'_u\times\vec r'_v}{|\vec r'_u\times\vec r'_v|}=\pm\frac{1}{|\vec r'_u\times\vec r'_v|}\left|\begin{array}\\\vec i&\vec j&\vec k\\x'_u&y'_u&z'_u\\x'_v&y'_v&z'_v\end{array}\right|$ , 易知方向余弦

\cos γ = \pm \frac{\frac{\partial (x, y)}{\partial (u, v)}}{| {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} |}

$\cos\gamma=\pm\frac{\frac{\partial(x,y)}{\partial(u,v)}}{|\vec r'_u\times\vec r'_v|}$ .
另一方面,

d S = | {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} | d u d v \Rightarrow P d x d y = P \cos γ d S = P [\pm \frac{\frac{\partial (x, y)}{\partial (u, v)}}{| {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} |}] | {\vec{r}}_{u}^{'} \times {\vec{r}}_{v}^{'} | d u d v = P (\pm \frac{\partial (x, y)}{\partial (u, v)}) d u d v

$\mathrm{d}S=|\vec r'_u\times\vec r'_v|\mathrm{d}u\mathrm{d}v\Rightarrow P\mathrm{d}x\mathrm{d}y=P\cos\gamma\mathrm{d}S=P\left[\pm\frac{\frac{\partial(x,y)}{\partial(u,v)}}{|\vec r'_u\times\vec r'_v|}\right]|\vec r'_u\times\vec r'_v|\mathrm{d}u\mathrm{d}v=P(\pm\frac{\partial(x,y)}{\partial(u,v)})\mathrm{d}u\mathrm{d}v$ .
同样可证明其余两式. ↩

54 先假定

L

$L$ 的形状满足这样的要求: 平行于

O y

$Oy$ 的直线至多与

L

$L$ 交于两点. 设下面曲线的方程为

y = y_{1} (x), a ⩽ x ⩽ b

$y=y_1(x),a\leqslant x\leqslant b$ , 上面曲线的方程为

y = y_{2} (x), a ⩽ x ⩽ b

$y=y_2(x),a\leqslant x\leqslant b$ .
于是

- \iint_{D} \frac{\partial P}{\partial y} d x d y = - \int_{a}^{b} d x \int_{y_{1} (x)}^{y_{2} (x)} \frac{\partial P}{\partial y} d y = - \int_{a}^{b} [P (x, y_{2} (x)) - P (x, y_{1} (x))] d x = ∳_{L} P d x

$-\iint\limits_D\frac{\partial P}{\partial y}\mathrm{d}x\mathrm{d}y=-\int_a^b\mathrm{d}x\int_{y_1(x)}^{y_2(x)}\frac{\partial P}{\partial y}\mathrm{d}y=-\int_a^b[P(x,y_2(x))-P(x,y_1(x))]\mathrm{d}x=\ointctrclockwise_LP\mathrm{d}x$ .
同样可以证明

\iint_{D} \frac{\partial Q}{\partial x} d x d y = ∳_{L} Q d y

$\iint\limits_D\frac{\partial Q}{\partial x}\mathrm{d}x\mathrm{d}y=\ointctrclockwise_LQ\mathrm{d}y$ . ↩

55 先假定

V

$V$ 是由上下两个显表示的曲面, 和周围是其母线平行于

O z

$Oz$ 的柱面所围成的.

V

$V$ 在

x O y

$xOy$ 上的投影为

D

$D$ . 设下面的曲面为

Σ_{1} : z = z_{1} (x, y), (x, y) \in D

$\Sigma_1:z=z_1(x,y),(x,y)\in D$ , 上面的曲面为

Σ_{2} : z = z_{2} (x, y), (x, y) \in D

$\Sigma_2:z=z_2(x,y),(x,y)\in D$ , 周围的柱面为

Σ_{3}

$\Sigma_3$ .
于是

\underset{V}{∭} \frac{\partial R}{\partial z} d x d y d z = \iint_{D} d x d y \int_{z_{1} (x, y)}^{z_{2} (x, y)} \frac{\partial R}{\partial z} d z = \iint_{D} [R (x, y, z_{2} (x, y)) - R (x, y, z_{1} (x, y))] d x d y = \iint_{{Σ_{2}}_{上}} R d x d y + \iint_{{Σ_{2}}_{下}} R d x d y

$\iiint\limits_V\frac{\partial R}{\partial z}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\iint\limits_D\mathrm{d}x\mathrm{d}y\int_{z_1(x,y)}^{z_2(x,y)}\frac{\partial R}{\partial z}\mathrm{d}z=\iint\limits_D[R(x,y,z_2(x,y))-R(x,y,z_1(x,y))]\mathrm{d}x\mathrm{d}y=\iint\limits_{{\Sigma_2}_\text{上}}R\mathrm{d}x\mathrm{d}y+\iint\limits_{{\Sigma_2}_\text{下}}R\mathrm{d}x\mathrm{d}y$ ,
而

\iint_{{Σ_{3}}_{外}} R d x d y = 0

$\iint\limits_{{\Sigma_3}_\text{外}}R\mathrm{d}x\mathrm{d}y=0$ , 所以

\iint_{V} \frac{\partial R}{\partial z} d x d y d z = \iint_{{Σ_{2}}_{上}} R d x d y + \iint_{{Σ_{2}}_{下}} R d x d y + \iint_{{Σ_{3}}_{外}} R d x d y = \underset{Σ_{外}}{∯} R d x d y

$\iint\limits_V\frac{\partial R}{\partial z}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\iint\limits_{{\Sigma_2}_\text{上}}R\mathrm{d}x\mathrm{d}y+\iint\limits_{{\Sigma_2}_\text{下}}R\mathrm{d}x\mathrm{d}y+\iint\limits_{{\Sigma_3}_\text{外}}R\mathrm{d}x\mathrm{d}y=\oiint\limits_{\Sigma_\text{外}}R\mathrm{d}x\mathrm{d}y$ .
同样对于

\iint_{V} \frac{\partial P}{\partial x} d x d y d z

$\iint\limits_V\frac{\partial P}{\partial x}\mathrm{d}x\mathrm{d}y\mathrm{d}z$ 及

\iint_{V} \frac{\partial Q}{\partial y} d x d y d z

$\iint\limits_V\frac{\partial Q}{\partial y}\mathrm{d}x\mathrm{d}y\mathrm{d}z$ 可得类似结果. ↩

56 由 Gauss 公式

\underset{Σ}{∯} \vec{v} \cdot \vec{n} d S = \underset{Σ_{外}}{∯} P d y d z + Q d z d x + R d x d y = ∭ (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) d V

$\oiint\limits_\Sigma\vec v\cdot\vec n\mathrm{d}S=\oiint\limits_{\Sigma_\text{外}}P\mathrm{d}y\mathrm{d}z+Q\mathrm{d}z\mathrm{d}x+R\mathrm{d}x\mathrm{d}y=\iiint(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z})\mathrm{d}V$ ,
易得. ↩

57 即要证

∳_{L} P d x + Q d y + R d z = \iint_{Σ +} (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) d y d z + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) d z d x + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y

$\ointctrclockwise_LP\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z=\iint\limits_{\Sigma+}(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})\mathrm{d}y\mathrm{d}z+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})\mathrm{d}z\mathrm{d}x+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y$ .
设

Σ : z = z (x, y), (x, y) \in D

$\Sigma:z=z(x,y),(x,y)\in D$ ,

D

$D$ 是

x O y

$xOy$ 上的闭区域, 边界为

L^{*}

$L^*$ .
设

Σ

$\Sigma$ 的边界

L

$L$ 的参数方程为

{\begin{aligned} x = x (t) \\ y = y (t) \\ z = z (x (t), y (t)) \end{aligned}, α ⩽ t ⩽ β

$\left\{\begin{split}&x=x(t)\\&y=y(t)\\&z=z(x(t),y(t))\end{split}\right.,\alpha\leqslant t\leqslant\beta$ , 故

L^{*}

$L^*$ 的参数方程显然就是

{\begin{array}{r} x = x (t) \\ y = y (t) \end{array}, α ⩽ t ⩽ β

$\left\{\begin{split}x=x(t)\\y=y(t)\end{split}\right.,\alpha\leqslant t\leqslant\beta$ .
于是

∳_{L} P (x, y, z) d x + Q (x, y, z) d y = ∳_{L^{*}} P (x, y, z (x, y)) d x + Q (x, y, z (x, y)) d y

$\ointctrclockwise_LP(x,y,z)\mathrm{d}x+Q(x,y,z)\mathrm{d}y=\ointctrclockwise_{L^*}P(x,y,z(x,y))\mathrm{d}x+Q(x,y,z(x,y))\mathrm{d}y$ ,
又

∳_{L} R d z = \int_{α}^{β} R \dot{z} d t = \int_{α}^{β} R (z_{x}^{'} \dot{x} + z_{y}^{'} \dot{y}) d t = ∳_{L^{*}} R z_{x}^{'} d x + R z_{y}^{'} d y

$\ointctrclockwise_LR\mathrm{d}z=\int_\alpha^\beta R\dot z\mathrm{d}t=\int_\alpha^\beta R(z'_x\dot x+z'_y\dot y)\mathrm{d}t=\ointctrclockwise_{L^*}Rz'_x\mathrm{d}x+Rz'_y\mathrm{d}y$ .
将以上两式相加, 应用 Green 公式, 得到

∳_{L} P d x + Q d y + R d z = ∳_{L^{*}} (P + R z_{x}^{'}) d x + (Q + R z_{y}^{'}) d y = \iint_{D} [\frac{\partial}{\partial x} (Q + R z_{y}^{'}) - \frac{\partial}{\partial y} (P + R z_{x}^{'})] d x d y

$\ointctrclockwise_LP\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z=\ointctrclockwise_{L^*}(P+Rz'_x)\mathrm{d}x+(Q+Rz'_y)\mathrm{d}y=\iint\limits_D[\frac{\partial}{\partial x}(Q+Rz'_y)-\frac{\partial}{\partial y}(P+Rz'_x)]\mathrm{d}x\mathrm{d}y$ ,
即

∳_{L} P d x + Q d y + R d z = \iint_{D} [(\frac{\partial Q}{\partial z} - \frac{\partial R}{\partial y}) z_{x}^{'} + (\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z}) z_{y}^{'} + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})] d x d y

$\ointctrclockwise_LP\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z=\iint\limits_D[(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y})z'_x+(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z})z'_y+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})]\mathrm{d}x\mathrm{d}y$ .
另一方面, 将

Σ : z = z (x, y)

$\Sigma:z=z(x,y)$ 视为参数为

x, y

$x,y$ 的参数方程, 应用参数曲面的第二种曲面积分的计算公式得到

\iint_{Σ +} (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) d y d z + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) d z d x + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y = \iint_{D} | \begin{matrix} \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} & \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} & \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \\ x_{x}^{'} & y_{x}^{'} & z_{x}^{'} \\ x_{y}^{'} & y_{y}^{'} & z_{y}^{'} \end{matrix} | d x d y

$\iint\limits_{\Sigma+}(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})\mathrm{d}y\mathrm{d}z+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})\mathrm{d}z\mathrm{d}x+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y=\iint\limits_D\left|\begin{array}\\\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}&\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}&\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\\x'_x&y'_x&z'_x\\x'_y&y'_y&z'_y\end{array}\right|\mathrm{d}x\mathrm{d}y$ ,
即

\iint_{Σ +} (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) d y d z + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) d z d x + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y = \iint_{D} [(\frac{\partial Q}{\partial z} - \frac{\partial R}{\partial y}) z_{x}^{'} + (\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z}) z_{y}^{'} + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})] d x d y

$\iint\limits_{\Sigma+}(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})\mathrm{d}y\mathrm{d}z+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})\mathrm{d}z\mathrm{d}x+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y=\iint\limits_D[(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y})z'_x+(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z})z'_y+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})]\mathrm{d}x\mathrm{d}y$ .
得证. ↩

58 旋度

rot \vec{v}

$\mathbf{rot}\,\vec v$ 有时也记成

curl \vec{v}

$\mathbf{curl}\, \vec v$ ↩

59 若取

\vec{n} = \cos α \vec{i} + \cos β \vec{j} + \cos γ \vec{k}

$\vec n=\cos\alpha\vec i+\cos\beta\vec j+\cos\gamma\vec k$ ,

L

$L$ 的内部为平面区域, 记作

Σ

$\Sigma$ ,
则

(rot \vec{v})_{M} \cdot \vec{n} = {| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} |}_{M} \cdot \vec{n} = {| \begin{matrix} \cos α & \cos β & \cos γ \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} |}_{M}

$(\mathbf{rot}\,\vec v)_M\cdot\vec n=\left|\begin{array}\\\vec i&\vec j&\vec k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\P&Q&R\end{array}\right|_M\cdot\vec n=\left|\begin{array}\\\cos\alpha&\cos\beta&\cos\gamma\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\P&Q&R\end{array}\right|_M$ , 右边明显是 Stokes 公式的形式, 易得. ↩

60 单连通性是不可缺少的,
例如

Σ : x^{2} + y^{2} > 0

$\Sigma:x^2+y^2>0$ , 积分曲线为

x^{2} + y^{2} = 1

$x^2+y^2=1$ ,
则

\oint \frac{x d y - y d x}{x^{2} + y^{2}} = \int_{0}^{2 π} (\cos^{2} t + \sin^{2} t) d t = 2 π \neq 0

$\oint\frac{x\mathrm{d}y-y\mathrm{d}x}{x^2+y^2}=\int_0^{2\pi}(\cos^2t+\sin^2t)\mathrm{d}t=2\pi\neq0$ .
但是却有

\frac{\partial}{\partial x} \frac{x}{x^{2} + y^{2}} = \frac{\partial}{\partial y} \frac{- y}{x^{2} + y^{2}}

$\frac{\partial}{\partial x}\frac{x}{x^2+y^2}=\frac{\partial}{\partial y}\frac{-y}{x^2+y^2}$ . ↩

61 易知原环流量与途径无关的充要条件是在

Σ

$\Sigma$ 内部作任一闭合曲线

l

$l$ , 皆有

\int_{l} P d x + Q d y

$\int_lP\mathrm{d}x+Q\mathrm{d}y$ .
由 Green 公式, 即证对任意域

D \in Σ

$D\in\Sigma$ 皆有

\iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y = 0

$\iint\limits_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y=0$ .
用反证法, 若存在某点使得

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} > 0

$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}>0$ , 则由

P, Q

$P,Q$ 偏微商的连续性, 存在包含该点的域

Σ_{0}

$\Sigma_0$ 使得

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} > 0

$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}>0$ .
易得. ↩

62 令

{\begin{aligned} P = x + y + 1 \\ Q = x - y^{2} + 3 \end{aligned}

$\left\{\begin{split}&P=x+y+1\\&Q=x-y^2+3\end{split}\right.$ , 因为满足确切微分条件, 所以一定有

u

$u$
使得

d u = (x + y + 1) d x + (x - y^{2} + 3) d y

$\mathrm{d}u=(x+y+1)\mathrm{d}x+(x-y^2+3)\mathrm{d}y$ .
因此, 一方面

d u = 0 \Rightarrow u = C

$\mathrm{d}u=0\Rightarrow u=C$ , 另一方面考虑选取途径为先

O x

$Ox$ 再

O y

$Oy$ 走向的折线段,
则

u (x, y) = \int_{(0, 0)}^{(x, y)} (x + y + 1) d x + (x - y^{2} + 3) d y = \int_{0}^{x} (x + 1) d x + \int_{0}^{y} (x - y^{2} + 3) d y = \frac{x^{2}}{2} + x + x y - \frac{y^{3}}{3} + 3 y

$u(x,y)=\int_{(0,0)}^{(x,y)}(x+y+1)\mathrm{d}x+(x-y^2+3)\mathrm{d}y=\int_0^x(x+1)\mathrm{d}x+\int_0^y(x-y^2+3)\mathrm{d}y=\frac{x^2}{2}+x+xy-\frac{y^3}{3}+3y$ .
因此, 方程的解为

3 x^{2} + 6 x + 6 x y - 2 y^{3} + 18 y = C

$3x^2+6x+6xy-2y^3+18y=C$ . ↩

63 设

λ, μ, ν

$\lambda,\mu,\nu$ 为任意三个外微分形式 (但

λ, μ

$\lambda,\mu$ 分别为

p, q

$p,q$ 次外微分形式), 则

d x \land d y = - d y \land d x

$\mathrm{d}x\wedge\mathrm{d}y=-\mathrm{d}y\wedge\mathrm{d}x$

(λ + μ) \land ν = λ \land ν + μ \land ν

$(\lambda+\mu)\wedge\nu=\lambda\wedge\nu+\mu\wedge\nu$

λ \land (μ + ν) = λ \land μ + λ \land ν

$\lambda\wedge(\mu+\nu)=\lambda\wedge\mu+\lambda\wedge\nu$

λ \land (μ \land ν) = (λ \land μ) \land ν

$\lambda\wedge(\mu\wedge\nu)=(\lambda\wedge\mu)\wedge\nu$

μ \land λ = (- 1)^{p q} λ \land μ

$\mu\wedge\lambda=(-1)^{pq}\lambda\wedge\mu$ ↩

64 此处仅讨论三维空间 ↩

ω = f

$\omega=f$ ↩

d ω = d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y + \frac{\partial f}{\partial z} d z \Rightarrow grad f = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k}

$\mathrm{d}\omega=\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z\Rightarrow\mathbf{grad}\,f=\frac{\partial f}{\partial x}\vec i+\frac{\partial f}{\partial y}\vec j+\frac{\partial f}{\partial z}\vec k$ ↩

ω_{1} = P d x + Q d y + R d z

$\omega_1=P\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z$ ↩

d ω_{1} = d P \land d x + d Q \land d y + d R \land d z = | \begin{matrix} d y \land d z & d z \land d x & d x \land d y \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} | \Rightarrow rot (P, Q, R) = | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} |

$\mathrm{d}\omega_1=\mathrm{d}P\wedge\mathrm{d}x+\mathrm{d}Q\wedge\mathrm{d}y+\mathrm{d}R\wedge\mathrm{d}z=\left|\begin{array}\\\mathrm{d}y\wedge\mathrm{d}z&\mathrm{d}z\wedge\mathrm{d}x&\mathrm{d}x\wedge\mathrm{d}y\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\P&Q&R\end{array}\right|\Rightarrow\mathbf{rot}\,(P,Q,R)=\left|\begin{array}\\\vec i&\vec j&\vec k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\P&Q&R\end{array}\right|$ ↩

ω_{2} = A d y \land d z + B d z \land d x + C d x \land d y

$\omega_2=A\mathrm{d}y\wedge\mathrm{d}z+B\mathrm{d}z\wedge\mathrm{d}x+C\mathrm{d}x\wedge\mathrm{d}y$ ↩

d ω_{2} = d A \land d y \land d z + d B \land d z \land d x + d C \land d x \land d y = (\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y} + \frac{\partial C}{\partial z}) d x \land d y \land d z \Rightarrow div (A, B, C) = \frac{\partial A}{\partial x} + \frac{\partial B}{\partial y} + \frac{\partial C}{\partial z}

$\mathrm{d}\omega_2=\mathrm{d}A\wedge\mathrm{d}y\wedge\mathrm{d}z+\mathrm{d}B\wedge\mathrm{d}z\wedge\mathrm{d}x+\mathrm{d}C\wedge\mathrm{d}x\wedge\mathrm{d}y=(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z})\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\Rightarrow\mathrm{div}\,(A,B,C)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ ↩